Download HW 1 Solutions

Homework 1 Solutions
1. (a) If x is an odd integer and y is an even integer, then xy is even.
(b) If x is an odd integer, then x2 is odd.
(c) If x is a prime number, then x2 is not prime.
2. (a)
(b)
(c)
(d)
(e)
At least one cow is not brown.
x > 2.
There is some cow that either doesn’t eat grass or isn’t brown.
x is not prime or x ≤ 12.
All cows either eat grass or are not black.
3. (a) Let P = “x > 1” and Q = “x > 0.” The statement “If x > 1, then x > 0” is true, but
“If x > 0, then x > 1,” is not true.
(b) “If ¬Q, then ¬P .” = “If x ≤ 0, then x ≤ 1.”
4. Let F be a field and x ∈ F . By the existence of inverses, we know that there is an element
−x ∈ F such that
x + (−x) = 0.
Similarly, the element −x has an additive inverse −(−x) such that
(−(−x)) + (−x) = 0.
Thus,
−(−x) + (−x) = x + (−x),
and we can cancel −x from both sides (as we showed in class) and end up with
−(−x) = x,
as desired.
5. Assume that
√
2 is rational. Then
√
a
,
b
for some a, b ∈ Z. We can assume that a and b have no common factors other than 1
(otherwise, we could reduce the fraction and get a new a and b).
So we have
√
2=
2b = a.
Squaring both sides yields
2b2 = a2 .
Thus, a2 is divisible by 2, so a is divisible by 2. By our assumption, this means that b is odd.
Since a is even, we know that a = 2c, where c ∈ Z. Thus,
a2 = 4c2 ,
so
2b2 = 4c2
=⇒ b2 = 2c2 .
Thus, b2 is even, so b is even, which is a contradiction! Thus,
√
2 must be irrational.
6. Let F be a field, and let a and b be elements of F . Suppose a · b = 0. Assume that a and b
are both nonzero (the negation of what we’re trying to show!). Then a has a multiplicative
inverse, so
a−1 (ab) = a−1 0.
By associativity and the proof in problem 4,
(a−1 a)b = 0
=⇒ b = 0.
This contradicts our assumption! Thus a = 0 or b = 0, as desired.
7. Let F be a field, and let a and b be nonzero elements of F . Since a is nonzero, it has a
multiplicative inverse, so
a · a−1 = 1.
Multiplying both sides by b yields
(a · a−1 )b = b
. By associativity,
a(a−1 b) = b.
Thus, such an element exists!
To show this element is unique, assume ac = b for some c. Then
ac = a(a−1 b)
so by the cancellation theorem in class,
c = a−1 b,
so such a c is indeed unique.
8. Let F be a field in R that contains Z. Let a/b be some rational number with a, b ∈ Z and b
nonzero. Then a, b ∈ F , since a and b are integers. So b has to have a multiplicative inverse
in F . Thus 1/b ∈ F . Since F is closed under multiplication,
a
1
=a
∈ F.
b
b
Thus, every element of Q is an element of F , so Q ⊆ F . So every field contained in R that
contains Z also contains Q. Thus, the minimal such field is Q itself.
9. Let F (along with the operations + and ·) be a field. Consider the set F 2 equipped with the
operations (a, b)+(c, d) = (a + c, b + d) and (a, b) ∗ (c, d) = (a · c, b · d). Assume that F 2 is a
field. Then for any (x, y) ∈ F 2 ,
(x, y) + (0, 0) = (x + 0, y + 0) = (x, y)
and
(x, y) ∗ (1, 1) = (x · 1, y · 1) = (x, y),
so (0, 0) must be the additive identity and (1, 1) the multiplicative identity. Thus (1, 0) is not
equal to the additive identity, so there must be some (a, b) ∈ F 2 such that
(1, 0) ∗ (a, b) = (1 · a, 0 · b) = (1, 1).
Thus, 0 = 0 · b = 1, which is a contradiction. Thus, F 2 is not a field.
10. Let X = {a, b, c, d}. Since X is a field (as we are told in the statement of the problem), we
know that it will have an additive identity and a multiplicative identity, so I’m just going to
set a = 0 and b = 1 right off the bat. Then our multiplication table looks like
·
0
1
c
d
0
0
0
0
0
1
0
1
c
d
c
0
c
d
0
d .
c is nonzero, so it has a multiplicative inverse. c · 0 = 0 and c · 1 = c, so c−1 is either c or d.
First assume c−1 = c. Then c2 = 1, and cd 6= 1 (by uniqueness of inverses). Thus, d has to
be its own multiplicative inverse as well, so d2 = 1.
But then what is cd? Well, it can’t be 0, since neither c nor d is 0. It isn’t 1, as we pointed
out already. It isn’t c since d 6= 1, and it isn’t d since c 6= 1. Thus, there are no remaining
possibilities, so our assumption that c2 = 1 must be incorrect.
Thus we know that c−1 = d.
What is c2 ? It is not 0 or 1, and it’s not c since c 6= 1. Thus, c2 = d. By analogous reasoning,
d2 = c, and we can fill in the rest of the multiplication table:
·
0
1
c
d
0
0
0
0
0
1
0
1
c
d
c
0
c
d
1
d
0
d .
1
c
Now, to fill in the addition table, we can start by finding additive inverses. Assume the
additive inverse of 1 is c. Then
c + d = c + c2 = c(1 + c) = c · 0 = 0,
which means that d is the additive inverse of c, which is a contradiction. Similarly, if d is the
additive inverse of 1, then
d + c = d + d2 = d(1 + d) = d · 0 = 0,
which is again a contradiction. Thus, it must be the case that 1 + 1 = 0, i.e. 1 is its own
additive inverse. Then by distributivity,
c + c = c(1 + 1) = c · 0 = 0
and
d + d = d(1 + 1) = 0.
We also know that 1 + c 6= 0, 1, or c (do you see why?), so 1 + c = d. Thus,
c + d = c(1 + c) = cd = 1,
and
1 + d = c(d + c) = c · 1 = c.
We can now fill in the addition table:
+
0
1
c
d
0
0
1
c
d
1
1
0
d
c
c
c
d
0
1
d
d
c .
1
0